36

Let's say I have one 2 kΩ resistor with 5% tolerance. If I replace it with two 1 kΩ resistors with 5% tolerance, will resulting tolerance go up, down, or remain unchanged?

I'm bad with probabilities, and I'm not sure what exactly tolerance says about resistance and its distribution.

I am aware that in the 'worst case' it will be the same; I'm more interested in what will happen on average. Will the chance of getting a more precise value increase if I use a series of resistors (because deviations will cancel each other out)?

On 'intuitive level' I think that it will, but I have no idea how to do the math with probabilities and find out if I'm actually right.

Peter Mortensen
  • 1,676
  • 3
  • 17
  • 23
Amomum
  • 675
  • 7
  • 14
  • Take the worst and best case of the tolerances and do the calculations for both scenarios yourself and compare. – PlasmaHH Apr 14 '16 at 11:37
  • 8
    This was a somewhat hotly contested issue a few years ago. See: [Reducing the tolerance of resistors manually](http://electronics.stackexchange.com/q/77325/25328) – Tut Apr 14 '16 at 11:41
  • 3
    \$2k\Omega 5\% = 2k\Omega\pm100\Omega \$ while \$1k\Omega 5\% = 1k\Omega\pm50\Omega \$, thus \$1k\Omega 5\%+1k\Omega 5\% = 2k\Omega\pm50\Omega \pm50\Omega = 2k\Omega\pm100\Omega\$ – Vladimir Cravero Apr 14 '16 at 11:43
  • @VladimirCravero that's the worse case, but what about the average? – Amomum Apr 14 '16 at 11:45
  • 3
    The average, as usual, is the nominal value. That's what nominal is there for. This assuming that R distribution is uniform in the tolerance range, which is not true. – Vladimir Cravero Apr 14 '16 at 11:48
  • 3
    Here's an interesting article that deals with the statistics, although the title is somewhat misleading if you accept tolerance as being worst-case: [Combining Multiple Resistors to Improve Tolerance](http://paulorenato.com/joomla/index.php?option=com_content&view=article&id=109&%E2%80%8C%E2%80%8BItemid=4) – Tut Apr 14 '16 at 12:50
  • @Tut that last link was exactly what I wanted to know! Can you please make an answer from your comment so I'll be able to accept it? – Amomum Apr 14 '16 at 13:35
  • Thanks, but ... link only answers are frowned upon and statistics are not in my area of expertise. Besides, I voted to close as a duplicate, and IMO there is a conflict of interest with voting to close and also providing an answer. I got the link from a comment in one of the answers to the question that I mentioned as a duplicate. That article needs to be taken with a grain of salt since the resistor values will likely not be random. – Tut Apr 14 '16 at 13:51
  • @Tut fair enough. – Amomum Apr 14 '16 at 13:59
  • 1
    It occurs to me that any "real" benefit or "debunked" reason is independent of what the circuit designer was thinking. Just because *we* know something is wrong doesn't mean the designer didn't act using that principle. So "should I do that" and "why does this board do that" are different questions. – JDługosz Apr 15 '16 at 09:07
  • Related, in depth video blog "posts" in the EEV blog: [one](https://www.youtube.com/watch?v=1WAhTdWErrU) and [two](https://www.youtube.com/watch?v=kSmiDzbVt_U) – try-catch-finally Apr 17 '16 at 07:26

8 Answers8

80

The worst case won't get any better. The result of your example is still 2 kΩ ±5%.

The probability that the result is closer to the middle gets better with multiple resistors, but only if each resistor is random within its range, which includes that it is independent of the others. This is not the case if they are from the same reel, or possibly even from the same manufacturer within some time window.

The manufacturer's selection process may also make the error non-random. For example, if they make resistors with a wide variance, then pick the ones that fall within 1% and sell them as 1% parts, then sell the remaining ones as 5% parts, the 5% parts will have a double-hump distribution with no values being within 1%.

Because you can't know the error distribution within the worst case error window, and because even if you did, the worst case stays the same, doing what you are suggesting is not useful to electronic design. If you specify 5% resistors, then the design must work correctly with any resistance within the ±5% range. If not, then you need to specify the resistance requirement more tightly.

Olin Lathrop
  • 310,974
  • 36
  • 428
  • 915
  • 6
    +1 for ... if each resistor has a random value **independently of the others** – Neil_UK Apr 14 '16 at 12:59
  • 7
    Excellent to point out that the manufacture may make different accuracies of the same resistor with the same process on the same line. This struck me as both disappointing, and completely sensible. – Dan Apr 14 '16 at 15:34
  • 2
    @Olin I'd even go a bit further about how the manufacturers "sort" the parts - they make a random batch of Rs, then they select as many "precision" valued (e.g. 1%) Rs as they need for the market expectation, and throw the rest to lower prec. ranges. The same goes with V tolerances for 1N400X diodes - I recall testing some DO-41 1N4001's just to realize they flawlessly worked for 230V AC... I asked a vendor about it, and he told me that *they have just a single production line* - they take as many 1N4003s as they need to from high-spec parts, *and sell all other as 1N4001* - YMMV, obviously. –  Apr 14 '16 at 16:14
  • @vaxquis Do you have any references regarding how manufacturers sort resistors (or define "tolerance")? When this was asked a few years ago, I looked at various manufacturers web-sites and datasheets and could find nothing. I emailed Vishay with several related questions, but got no response. – Tut Apr 14 '16 at 17:18
  • 6
    @Tut: I doubt manufacturers are going to tell you how they test and sort parts. All they are going to say is that 5% parts will be within 5% of the nominal value, and that's all you should care about anyway. Strategies for binning parts can change. If it's not in the datasheet, then don't count on it and don't try to guess or assume beyond it. – Olin Lathrop Apr 14 '16 at 17:28
  • @OlinLathrop I think you are probably right which explains why I was unable to get an answer (although I think they should at least tell you whether or not they test 100% of the parts). There seems to be a lot of speculation about this (or people claiming to know how they do it) and even debates on the definition of "tolerance" as used by the manufacturers. The reason I didn't answer a few years ago was that I preferred not to speculate. – Tut Apr 14 '16 at 17:39
  • 2
    @Tut https://www.maximintegrated.com/en/app-notes/index.mvp/id/5663 `We say "seems to" and "appears to" because sales volume and human nature also influence the mix. For example, the plant manager may need to ship 5% tolerance capacitors, but he does not have enough to meet the demand this month. He does, however, have an overabundance of 2% tolerance parts. So, this month he throws them into the 5% bin and makes the shipment. Clearly deliberate, human intervention can, and does, skew the statistics and method.` –  Apr 14 '16 at 18:47
  • 1
    @Tut BTW you can check it for yourself - buy some Rs or diodes, and check their characteristics. As I said, often (in my experience, more often than not, when dealing with some vendors), you'll find that "5%" Rs are actually within +/- 1%-2% of the spec, that you can pump *excessive* Vs through diodes with lower voltage rating for extended time without any failures etc. It's similar to how core locking was used in https://en.wikipedia.org/wiki/Phenom_II - sometimes the disabled cores were indeed faulty, sometimes they were 100% funcional. It's all about sale quotas IMO. –  Apr 14 '16 at 18:53
  • 1
    Unless we posit superconductors connecting the resistors to each other, the trace or writing between them will additionally have *some* resistance of its own. *Miniscule, yes, but non-zero.* So the maximum resistance of two 1 kOhm +/- 5% resistors (for a true, exact 5% maximum error) in series would seem to be ever so slightly above 2 kOhm plus 5%, and the minimum would seem to be ever so slightly above 2 kOhm minus 5%. I agree with the point of your answer, though; if specified +/- 5%, then the design must work correctly with anything within the +/- 5% range. – user Apr 15 '16 at 13:39
  • Obviously that was supposed to be "the trace **or wiring**". Stupid keyboard. (And using superconductors to connect two resistors in series seems... silly. To put it mildly.) – user Apr 15 '16 at 22:43
  • 1
    For some real data, including from the same batch of resistors and from different batches, see *[Gaussian Resistors](https://www.eevblog.com/2011/11/08/eevblog-215-gaussian-resistors/)* (32 min 48 secs) and *[Gaussian Resistor Redux](https://www.eevblog.com/2011/11/14/eevblog-216-gaussian-resistor-redux/)* (34 min 39 secs). – Peter Mortensen Apr 16 '16 at 12:38
  • 1
    *war story* Supplied a small manufacturer with 500 kits for a pilot run from our normal stock expecting 1% dropout rate. Result:0% dropout. Customer used their own (cheap) supplier - 33% dropouts. On bench test, power transistor gain – Magoo Apr 17 '16 at 08:46
8

The answer depends a lot on the distribution of the real resistor values, and what your question actually is.

I did a simulation, for which I generated a set of 100,000 resistors with 1% tolerance (easier to handle than 5%). From this, I took 1,000,000 times a sample of two and calculated the sum of them.

For the set, I assumed three different distributions:

  1. A narrow, perfectly gaussian distribution with \$\sigma=2.5\$. This means: 63% of all resistors are in the range \$1000\pm2.5\Omega\$ and 99.999998% are in the range \$1000\pm10\Omega\$.
    Think of a manufacturer with a reliable production process here. If he wants 1kOhm resistors with 1%, his machine produces them.

  2. A uniform distribution where the probability to get any value in the 1% range is equal.
    Think of a manufacturer with a very unreliable production process. The machine produces resistors of any value of a wide range, and he has to pick out the 1%/1kOhm resistors.

  3. A wide gaussian distribution (\$\sigma=5\$), where every resistor outside the 1% range is thrown away and replaced by a "good" one. This is just a blend of the first two cases.
    This is a manufacturer with a better process. Most of the resistors meet the specs, but some have to be sorted out.

Here is the result:

enter image description here

  1. When adding two values of the same gaussian distribution, the sum also is a gaussian distribution with a width of \$\sigma_{new}=\sqrt{2}\sigma_{old}\$.
    The resistors have a tolerance of \$\pm 10\Omega\$, which converts to a new tolerance of \$\pm 14.1\omega\$ or \$14.1\Omega/2000\Omega=0.7\%\$.
    The simulated data shows this, too, as the distribution is slightly wider that 0.5% (vertical green lines)

  2. The uniform distribution becomes a triangular distribution. You still get resistor pairs of 1980 or 2020 Ohms (5%), but there are more combinations with lower difference from the nominal value.

  3. The result also is a blend of the results of the first two cases...

As said in the beginning, it depends on the distribution. In any case, the probability is higher to get a resistance with less difference from the nominal value, but there's still a probability to get a value which is 1% off.

Further notes:

  • Often, a batch contains resistors which all have nearly the same value, which is a bit off the nominal value. E.g. they are all in the range of 995...997Ohm, which is still well in the range of 990...1010Ohm. By combining two resistors, you get a lower spread, but the values are all a little low.

  • Resistors show e.g. temperature dependence. The precision is much better than 1% to ensure the resistance stays in the 1% range at different temperatures.

sweber
  • 8,580
  • 1
  • 15
  • 35
  • 3
    Unfortunately, your thought experiment is mostly disqualified by that "further note" - the error cannot be expected to be random, rather it will probably have a consist bias, or else a few consistent biases if your pool contains multiple manufacturing lots. – Chris Stratton Apr 14 '16 at 16:11
  • 2
    Also if you take 5% resistor built by selecting good enough "failed" resistors from a 1% manufacturing line then the distribution will be off even more. – ratchet freak Apr 15 '16 at 11:55
  • Your graphs use "norm" as a label for the uniform distribution. ["Normal distribution"](https://en.wikipedia.org/wiki/Normal_distribution) is another term for "Gaussian distribution", so it's very a poor choice. – Peter Cordes Apr 17 '16 at 01:33
  • @PeterCordes: Absolutely right, fixed! – sweber Apr 18 '16 at 07:22
3

Fun question, Practically, when I was looking at 1% 1/4 W Metal Film R's I found that in a batch, the distribution was far from random. Most of the R's clustered around a value that could be a bit above or a bit below the "target" value. So at least for the R's I looked at it wouldn't make any difference.

George Herold
  • 4,754
  • 1
  • 18
  • 25
1

There are two important numbers that have to do with your question.

The first is "Worst Case Scenario": In the absolute worst case, one 2k resistor with 5% will be either 2.1k or 1.9k. One resistor of 1k 5% will be 1.05k or 0.95k, added together this comes to either 2.1k or 1.9k. So in the worst case, in series, a bunch of resistors with the same tollerance will always retain their tollerance over the total value and be just as good as one big one.

The other important number is the law of large numbers. If you have 1000 resistors that have an ideal target value and are specified with an absolute maximum error of 5%, of course it's very likely that quite a few of those will be very close to the target value and that the number of resistors with too high a value is about as high as the number with a lower value. The production process for components like resistors falls under a natural statistical process, so it's extremely likely the resulting resistors in a large batch across multiple productions yield what is called a gaussian curve. Such a curve is symetrical around the "desired" value and the manufacturer will try to get that "desired" value to be the value he sells the resistors as, for statistical yield reasons. So you can make an assumption that if you buy 100 resistors, you too get a gaussian distribution. Actually, that may not be the exact case, with resistors a large enough number may have to be 10's of thousands to get a real gaussian distribution. But the assumption is more valid than that all will be off by the worst case in the same direction (all with -5%, or all with +5%)

That's all well and nice, but what does it mean? It means that if you have 10 resistors of 200 Ohms at 5% in series, it's reasonably likely that one will be 201 Ohm, another 199 Ohm, another will be 204 Ohm, yet another will be 191 Ohm, etc etc, and all those "too low" and "too high" values compensate each other and it becomes, suddenly, a big 2k chain with a much better accuracy, through the law of large numbers.

Again, this is only in the specific case of the same value resistors in series. While different values in series are also likely to become more accurate on average, the degree to which this happens or how likely it is, is hard to express correctly without knowing the exact use-case and exact-values.

So, it is, at the least, not at all harmful to place many resistors of same value in series, and usually it gives a much better result. Combine that with the fact that manufacturing a huge amount of boards with just 3 different components is much cheaper than with 30 different components and you often see designs with only 1k and 10k (or maybe 100 Ohm and 100k as well)resistos in cheap, high-volume-production trinkets, where any other value is a combination of the two.

Asmyldof
  • 18,299
  • 2
  • 33
  • 53
  • 1
    Even tens of thousands may not be enough to ensure you get resistors from different batches. Resistor production is something that happens on a massive scale. – Peter Green Apr 14 '16 at 15:08
  • @PeterGreen True. But, from experience I can say that at least Yageo and TE have within-batch differentiation that is well measurable across even a 10piece length of strip. Where any variation within the tolerance band guarantees better than tolerance end value. That said the variation across a 100unit strip often proofs to be less than 1/4th tolerance and usually not balanced around the target value. – Asmyldof Apr 14 '16 at 16:19
0

In terms of the maximum/minimum prossible deviation, both cases present the same result.

If you consider the probability of ocurring a 1% deviation to be the same of a 5% deviation, then both cases present the same result.

If you consider the deviation to follow some sort of normal distribution, centered on the resistor design value, still makes no diference. Because even thought the individual deviations will be smaller, the sum will bring them close to the deviations of a bigger resistor. The probability of a 0.5% deviation in a 2kOhm resistor is the same as in a 1kOhm resistor, even though the value of the deviation differs.

AmiguelS
  • 275
  • 1
  • 8
  • 1
    If the resistors indepently followed a normal distribution then using multiple resitors would be an improvement. The problem is that resistors don't tend to do that, there is a very high correlation in value between multiple resistors from the same batch and chances are if you order a bunch of resistors of the same nominal value they will all have come from the same batch. – Peter Green Apr 14 '16 at 15:10
0

Solid carbon resistors have all but ceased to exist on the market as they catch fire easily and change value with voltage. Now days 'carbon' is normally carbon film.

It is a much more stable resistor, but not as stable as metal-film or ultra-stable like the ceramic resistors made by Caddock. Usually 0.025% is available for about $50 each. A laboratory grade 0.01% or better cost about $150-for now.

Most boards that I work with use 1% metal film smd, which now have a very low cost after being on the market for several decades. Stability with temperature and time is often more important than the absolute value of the resistor.

I sometimes put in a notice in the users guide for my test equipment, to turn it on 15 minutes early so the readings for voltage or current are within 0.1% worst case. If I have to I manually pick series or parallel resistors for absolute value, from a batch that is stable enough over time (10 - 20 years) to be useful in production.

I do not use trim-pots unless mandatory, as their drift is about 200 ppm. If I have to use a trim-pot I use series resistors to keep the trim-pot value low as possible.

For 'surge' resistors I usually had to use 14 awg nickel-chrome wire, 30 strands in parallel to handle 10,000 to 150,000 amp surges of about 20 uS duration each. Exact resistive values were not as important as survivability.

In this sense they were much like wirewound resistors on steroids. The accuracy was seldom better than 10% and they drifted with temperature several percent. They ran too hot to touch, but this was normal, it was about surviving a harsh environment.

We used 6awg wire inductors in series with 0.1 ohm ceramic donut resistors rated for 10,000 amp surges for wave-shaping. Connections were made with buss-bars or 500 mcm locomotive cable. The 'emergency dump' is a water tower resistor made with water and copper sulfate, 3 inch's diamater and about a meter in height. It had a resistance of about 500 ohms but was the only resistor that could dump the charge (30,000 volts) without blowing up.

You can split hairs all you want over deviation, but in the end you build with what works. Sometimes tolerance has to take a backseat to other issues.

I have seen deviation in precision resistors, say reels of 5,000, that seem to drift above or below the ideal value (as measured by a Fluke 87 DVM). It makes finding a series / parallel combination with an exact values nearly impossible. I simply use those that have the closest 'fit' to the value needed.

At ultra-precision levels (<0.025%), controlling temperature drift,board leakage and noise becomes a big issue. Now you have to add parts to keep the 'deviation' over time from becoming an issue.

In terms of measuring with precision equipment (0.01% or better), no combination of series or parallel resistors can be more accurate over time than one resistor that already has a deviation so close to zero as to not be an issue.

Multiple resistors in series or parallel create multiple instances of temperature drift and deviation. To expect them to 'null' out the deviations is absurd, because temperature drift is always an 'additive' function, and deviations tend to drift in one direction on reels of 5,000, yet meet the tolerance spec.

To create a 'perfect' resistor value from multiple values, those with positive deviation would need a negative temperature coefficient, while those in series or parallel that have a negative deviation would need a positive temperature coefficient. Both types of coefficients would have to match to cancel out temperature drift.

From my point of view, during practical normal usage, my answer to @Amomum is NO.

-1

The probabilty is $$E_{sum}=\frac{1}{N} \sqrt{E_{1}^{2} + E_{2}^{2}+ .. +E_{N}^{2} }$$ so $$E_{sum}=\frac{1}{2}\sqrt{5^{2}+5^{2}}= 3.53% $$ enter image description here

The image of tolerance shows how resitors are sorted during the production process. They are distributes in bins containing specified tolerance, so for example in the bin containing +/-10% you won't find any resistor that has better tolerance than >+/-5%, because those parts are contined in the bin of +/-5%. But if you biuld a series chain a large number of resistors, the mean value will be close to specified $$R=nR$$.

Marko Buršič
  • 23,562
  • 2
  • 20
  • 33
  • 2
    You got downvoted because there is no expectation of randomness in a batch of resistors. – Scott Seidman Apr 14 '16 at 13:38
  • @ScottSeidman According to others answers there is a neat expectation of randomness. Why then the elements are marked +/- % if there is no randomness? – Marko Buršič Apr 14 '16 at 14:39
  • 2
    The components have a tolerance for deviation from their nominal value. But the *distribution* of the error cannot be expected to be *random*. In fact it is quite unlikely to be. The mathematical concept of "random" (on which your calculation depends) has a far more specific meaning than "unknown" which is the actual situation. – Chris Stratton Apr 14 '16 at 16:09
  • It is not random, it is Gaussian distribution. In the production process the machine is adjusted so the mean value of the resistance is the specified resistance, the parts are then sorted into the bins with specified tolerance. Let we have 3 bins +/-1%, +/-5%, +/-10%. In the first there are resistances all around specified value, in the second bin you won't find any resitor within +/-1% tolerance, but the mean value is still the same. Same for +/-10% bin. Within the same lot of produced resistances the average resistance of the entire lot is specified resistance. – Marko Buršič Apr 14 '16 at 16:36
  • 3
    @MarkoBursic Do you get this information from some sort of research / experience or just intuition? If the latter, the reality might be different as more precise resistors are usually made with a different process completely. – akaltar Apr 14 '16 at 17:07
  • @akaltar OK, then I expect you will post your answer. I am looking forward for your explanation. – Marko Buršič Apr 14 '16 at 17:29
  • 1
    @MarkoBursic I don't want to be mean here. I don't know the correct answer to this question. I just usually see that 1% resistors are "Metal film" while usually 5% resistors are "carbon", so I assume they are usually made differently. I just wanted to know if this is actually insider information, in which case I am wrong. Tough assuming this distribution is the actual one, your answer is good. – akaltar Apr 14 '16 at 17:59
  • 1
    It probably is a Gaussian distribution of error -- most things are. What I mean to say is that the distribution of error is very likely NOT to have a zero mean. In other words, the mean resistance is not likely to be the nominal value – Scott Seidman Apr 14 '16 at 20:18
-2

Tolerance means the limit over which the value may get diverge from its actual value. 5% 2k resistor means that the resistance will have value between 1900ohm to 2100ohms. Now for two 1k resistors the value of tolerance will add upand becomes 10%. This a simple rule of Errors. You can read more about this in any Instrumentation and Measurement book. So this means that the value two 1k resistor will vary between between 1800ohm to 2200ohms.

Prashant Joshi
  • 7
  • 2
  • 2
    Just plain wrong. Two 1 kOhm 5% resistors in series don't make a 2 kOhm 10% resistor. The tolerances do not add like that. – Olin Lathrop Apr 17 '16 at 13:23